Question

How do you evaluate ${}^8{C_2}$?

Answer 1

In the above question you were asked to solve ${}^8{C_2}$. This is a problem of combination. The combination is of two types: one in which repetition is allowed and second without repetition. The formula of combination is ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ , which you can use to solve this question. We will also use the factorial formula in the solution. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given problem, we have to evaluate ${}^8{C_2}$. Here, we will use the formula of combination which is $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$. You can easily identify that n is 8 and r is 2 here.
${ \Rightarrow ^8}{C_2} = \dfrac{{8!}}{{(8 - 2)!.2!}}$
$= \dfrac{{8!}}{{6!.2!}}$
$= \dfrac{{8 \times 7 \times 6!}}{{6!.2!}}$
After calculating the above e\times pression, we get
$= \dfrac{{8 \times 7}}{{2!}}$
$= \dfrac{{8 \times 7}}{{2 \times 1}}$
$= \dfrac{{56}}{2}$
$= 28$
Therefore, ${}^8{C_2} = 28$.

Note:
In the above solution we have used factorials. The formula of factorial for n is, $n! = n \times (n-1) \times (n-2) \times (n-3)…. 3 \times 2 \times 1$. Also, we used the combination’s formula which is ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ to solve the problem. The formula of combination is used to the total number of ways in which the items can be selected. In other words, the selection of objects from a larger group is called combination but the order of selection does not matter.